Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $p = \dfrac{5z^2 + 5z - 100}{-5z^2 + 15z + 20} \div \dfrac{-4z^2 - 20z}{2z^2 + 16z} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{5z^2 + 5z - 100}{-5z^2 + 15z + 20} \times \dfrac{2z^2 + 16z}{-4z^2 - 20z} $ First factor out any common factors. $p = \dfrac{5(z^2 + z - 20)}{-5(z^2 - 3z - 4)} \times \dfrac{2z(z + 8)}{-4z(z + 5)} $ Then factor the quadratic expressions. $p = \dfrac {5(z - 4)(z + 5)} {-5(z - 4)(z + 1)} \times \dfrac {2z(z + 8)} {-4z(z + 5)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { 5(z - 4)(z + 5) \times 2z(z + 8)} { -5(z - 4)(z + 1) \times -4z(z + 5)} $ $p = \dfrac {10z(z - 4)(z + 5)(z + 8)} {20z(z - 4)(z + 1)(z + 5)} $ Notice that $(z - 4)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {10z\cancel{(z - 4)}(z + 5)(z + 8)} {20z\cancel{(z - 4)}(z + 1)(z + 5)} $ We are dividing by $z - 4$ , so $z - 4 \neq 0$ Therefore, $z \neq 4$ $p = \dfrac {10z\cancel{(z - 4)}\cancel{(z + 5)}(z + 8)} {20z\cancel{(z - 4)}(z + 1)\cancel{(z + 5)}} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $p = \dfrac {10z(z + 8)} {20z(z + 1)} $ $ p = \dfrac{z + 8}{2(z + 1)}; z \neq 4; z \neq -5 $